= Solution
{c}
Reduction modulo the prime $p$ preserves integer addition, multiplication, and one:
$$
\overline{a+b}=\bar a+\bar b,\qquad \overline{ab}=\bar a\bar b.
$$
Every congruence class has an integer representative, so the <ring homomorphism> is onto. Its kernel is exactly $p\mathbb Z=(p)$. The quotient is a field because every class not divisible by $p$ has a multiplicative inverse by <Bézout's identity>.
For the coefficientwise extension to the <polynomial ring>, addition is preserved term by term. If $f=\sum_ia_ix^i$ and $g=\sum_jb_jx^j$, the coefficient of $x^k$ in their product is $\sum_{i+j=k}a_ib_j$. Reducing it modulo $p$ gives $\sum_{i+j=k}\bar a_i\bar b_j$, the coefficient in $\hat\phi(f)\hat\phi(g)$. Thus multiplication and one are preserved as well. Lift each coefficient of a polynomial over $\mathbb Z_p$ to an integer to prove surjectivity. The kernel is
$$
\boxed{\ker\hat\phi=p\mathbb Z[x]=(p).}
$$
Indeed, its elements are precisely the polynomials whose coefficients are all divisible by $p$.
In the <binomial theorem> expansion of $(x-1)^p$, every intermediate coefficient $\binom pj$ is divisible by $p$: the numerator contains the prime factor $p$, while $j!(p-j)!$ does not for $1\leq j<p$. Hence
$$
(x-1)^p=x^p+(-1)^p=x^p-1\quad\text{in }\mathbb Z_p[x].
$$
For $p=2$, the last equality uses $1=-1$ in characteristic two. This is an instance of the <Frobenius endomorphism>.
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